120
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
SOLUTIONS TO SELECTED TEXT PROBLEMS
7.7
7.8
(a)
λ=
(b)
ν=
3.00 × 108 m / s
13
8.6 × 10
/s
3.00 × 108 m / s
5.66 × 10−7 m
= 3.5 × 10−6 m = 3.5 × 103 nm
= 5.30 × 1014 / s = 5.30 × 1014 Hz
(a)
Step 1: c = λν. Solve this equation algebraically for the frequency (ν).
ν =
c
(7.1)
λ
Step 2: Since the speed of light has units of m/s, we must convert the wavelength from units of nm to m.
456 nm ×
1 × 10 −9 m
−7
= 4.56 × 10 m
1 nm
Step 3: Calculate the value of ν by substituting the known quantities into Equation (7.1).
m
s = 6.58 × 1014 s−1 = 6.58 × 1014 Hz
ν =
−7
4.56 × 10 m
3.00 × 108
(b)
Step 1: c = λν. Solve this equation algebraically for the wavelength (λ).
λ =
c
(7.2)
ν
Step 2: Calculate the value of λ by substituting in the known quantities into Equation (7.2).
m
s = 0.122 m
λ =
9 1
2.45 × 10
s
3.00 × 108
Step 3: Convert the wavelength in units of meters to nanometers.
λ = 0.122 m ×
7.9
1 nm
1 × 10 −9 m
8
= 1.22 × 10 nm
8
Since the speed of light is 3.00 × 10 m/s, we can write

 1.61 km  1000 m  
1s
1.3 × 108 mi 
 = 7.0 × 102 s

 
8
1
mi
1
km


  3.00 × 10 m 
Would the time be different for other types of electromagnetic radiation?
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
7.10
time =
121
distance
speed
A radio wave is an electromagnetic wave, which travels at the speed of light. The speed of light is in units of
m/s, so let’s convert distance from units of miles to meters.
7
? distance (m) = (2.8 × 10 mi) ×
time =
7.11
λ=
161
. km
1000 m
10
= 4.5 × 10 m
×
1 mi
1 km
4.5 × 1010 m
distance
2
= 1.5 × 10 s = 2.5 min
=
8
speed
3.00 × 10 m / s
c
3.00 × 108 m / s
=
= 3.26 × 10−2 m = 3.26 × 107 nm
1
−
ν 9192631770 s
This radiation falls in the microwave region of the spectrum.
7.12
The wavelength is:
λ =
ν =
1m
−7
= 6.057802106 × 10 m
1,650,763.73 wavelengths
c
λ
=
3.00 × 108 m / s
6.057802106 × 10 −7 m
= 4.95 × 10
14 −1
s
hc (6.63 × 10−34 J s)(3.00 × 108 m / s)
=
= 3.19 × 10−19 J
−9
λ
624 × 10 m
7.15
E = hν =
7.16
(a)
Step 1: c = λν. Solve this equation algebraically for the wavelength (λ).
λ =
c
(7.2)
ν
Step 2: Calculate the value of λ by substituting the known quantities into Equation (7.2).
λ =
(b)
m
s = 4.0 × 10−7 m = 4.0 × 102 nm
14 1
7.5 × 10
s
3.00 × 108
E = hν. Substitute the frequency (ν) into this equation to solve for the energy of a single photon
associated with this frequency.
E = hν = (6.63 × 10
7.17
(a)
λ=
−34
J⋅s)(7.5 × 10
14 1
s
) = 5.0 × 10
c 3.00 × 108 m / s
=
= 5.0 × 103 m = 5.0 × 1012 nm
ν
6.0 × 104 / s
The radiation does not fall in the visible region; it is radio radiation.
−19
J
122
7.18
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
−34
(b)
E = hν = (6.63 × 10
(c)
Converting to J/mol:
4
J⋅s)(6.0 × 10 /s) = 4.0 × 10
E=
−29
J
4.0 × 10−29 J
6.022 × 1023 photons
−5
= 2.4 × 10 J/mol
×
1 photon
1 mol
The energy given in this problem is for 1 mole of photons. To apply E = hν, we must divide by Avogadro’s
number. The energy of one photon is:
E =
10
. × 103 kJ
1 mol
1000 J
−18
= 1.7 × 10 J/photon
×
×
23
1 mol
1
kJ
6.022 × 10 photons
The wavelength of this photon can be found using the relationship E = hc/λ.
hc
=
λ =
E
m

(6.63 × 10−34 J ⋅ s)  3.00 × 108 
 1 nm 
s 
2

× 
 = 1.2 × 10 nm
−18
 1 × 10−9 m 
1.7 × 10
J


The radiation is in the ultraviolet region (see Figure 7.2 of the text).
7.19
E = hν =
7.20
(a)
λ =
hc ( 6.63 × 10 −34 J ⋅ s )(3.00 × 108 m / s )
=
= 1.29 × 10 −15 J
λ
( 0154
.
× 10 −9 m )
c
ν
m
s = 3.70 × 10−7 m = 3.70 × 102 nm
λ =
14 1
811
. × 10
s
3.00 × 108
(b)
Checking Figure 7.2 of the text, you should find that the visible region of the spectrum runs from 400 to
700 nm. 370 nm is in the ultraviolet region of the spectrum.
(c)
E = hν. Substitute the frequency (ν) into this equation to solve for the energy of one quantum
associated with this frequency.
E = hν = (6.63 × 10
−34
J⋅s)(8.11 × 10
14 1
s
) = 5.38 × 10
−19
J
7.25
The arrangement of energy levels for each element is unique. The frequencies of light emitted by an element
are characteristic of that element. Even the frequencies emitted by isotopes of the same element are very
slightly different.
7.26
The emitted light could be analyzed by passing it through a prism.
7.27
Light emitted by fluorescent materials always has lower energy than the light striking the fluorescent
substance. Absorption of visible light could not give rise to emitted ultraviolet light because the latter has
higher energy.
The reverse process, ultraviolet light producing visible light by fluorescence, is very common. Certain
brands of laundry detergents contain materials called “optical brightners” which, for example, can make a
white shirt look much whiter and brighter than a similar shirt washed in ordinary detergent.
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
7.28
Excited atoms of the chemical elements emit the same characteristic frequencies or lines in a terrestrial
laboratory, in the sun, or in a star many light-years distant from earth.
7.29
(a)
The energy difference between states E1 and E4 is:
E4 − E1 = (−1.0 × 10
λ=
(b)
−19
)J − (−15 × 10
−19
)J = 14 × 10
−19
J
hc ( 6.63 × 10 −34 J ⋅ s )(3.00 × 108 m / s )
=
= 1.4 × 10 −7 m or 1.4 × 102 nm
∆E
14 × 10 −19 J
The energy difference between the states E2 and E3:
E3 − E2 = (−5.0 × 10
(c)
123
−19
J) − (−10.0 × 10
−19
J) = 5 × 10
−19
J
The energy difference between the states E1 and E3 is:
E1 − E3 = (−15 × 10
−19
J) − (−5.0 × 10
−19
J) = −10 × 10
−19
J
Ignoring the negative sign of ∆E, the wavelength is found as in part (a)
λ=
7.30
hc ( 6.63 × 10 −34 J ⋅ s )(3.00 × 108 m / s )
=
= 2.0 × 10 −7 m or 2.0 × 102 nm
∆E
10 × 10 −19 J
We use more accurate values of h and c for this problem.
hc
E = hν =
7.31
λ
(6.6256 × 10 −34 J ⋅ s)(2.998 × 108 m / s)
=
656.3 × 10
−9
m
= 3.027 × 10
−19
J
In this problem ni = 5 and nf = 3.
∆E = RH
F 1 − 1 I = (218
GH n n JK . × 10
i
2
f
−18
2
FG 1 − 1 IJ = −155
H 5 3 K . × 10
J)
2
2
−19
J
The sign of ∆E means that this is energy associated with an emission process.
λ=
hc ( 6.63 × 10 −34 J ⋅ s )(3.00 × 108 m / s )
=
= 1.28 × 10 −6 m or 1.28 × 103 nm
−19
∆E
155
. × 10
J
Is the sign of the energy change consistent with the sign conventions for exo− and endothermic processes?
7.32
Step 1: Using the following equation, we can calculate the energy change for the transition.
∆E = RH
F 1 − 1I
GH n n JK
2
i
∆E = (2.18 × 10
2
f
−18
∆E = −4.09 × 10
−19
J)
J
F 1 1I
GG − JJ
H4 2 K
2
2
124
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
Step 2: The negative sign for ∆E indicates that energy is released to the surroundings during the emission
process. To calculate the wavelength, we omit the minus sign for ∆E because the frequency of the
photon must be positive. We know that
∆E = hν
Rearranging the equation and substituting in the known values,
ν =
∆E
(4.09 × 10−19 J)
14 −1
=
= 6.17 × 10 s or Hz
−34
h
(6.63 × 10
J ⋅ s)
Step 3: We also know that λ =
c
ν
. Substituting the frequency calculated above into this equation gives:
FG 3.00 × 10 m IJ
H
sK
7
λ =
= 4.86 × 10 m = 486 nm
1I
FG 617
H . × 10 sJK
8
−
14
7.33
This problem must be worked to four−significant−figure accuracy. We use 6.6256 × 10
constant. First calculate the energy of each of the photons.
E=
hc ( 6.6256 × 10 −34 J ⋅ s )( 2.998 × 108 m / s )
=
= 3372
× 10 −19 J
.
λ
589.0 × 10 −9 m
E=
hc ( 6.6256 × 10 −34 J ⋅ s )( 2.998 × 108 m / s )
=
= 3369
× 10 −19 J
.
−9
λ
589.6 × 10 m
−34
J⋅s for Planck’s
For one photon the energy difference is:
∆E = (3.372 × 10
−19
J) − (3.369 × 10
−19
J) = 3 × 10
−22
J
For one mole of photons the energy difference is:
F 3 × 10 J I F 6.022 × 10 photons I = 2 × 10 J / mol
GH 1 photon JK GH
JK
1 mol
−22
7.34
∆E = RH
23
2
F 1 − 1I
GH n n JK
2
i
2
f
nf is given in the problem and RH is a constant, but we need to calculate ∆E. The photon energy is:
E =
hc
λ
=
(6.63 × 10 −34 J ⋅ s)(3.00 × 108 m / s)
434 × 10
−9
m
= 4.58 × 10
−19
J
Since this is an emission process, the energy change ∆E must be negative, or −4.58 × 10
−19
J.
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
Substitute ∆E into the following equation, and solve for ni.
∆E = RH
F 1 − 1I
GH n n JK
−4.58 × 10
2
i
2
f
−19
J = (2.18 × 10
−18
F 1 − 1I
GH n 2 JK
J)
i
1
ni 2
2
2
F −4.58 × 10− J I + 1 = −0.210 + 0.250 = 0.040
GH 2.18 × 10− J JK 2
19
=
ni =
18
1
0.040
2
= 5
h
6.63 × 10 −34 J ⋅ s
=
= 5.65 × 10 −10 m = 0.565 nm
mu
(1675
.
× 10 −27 kg )( 7.00 × 102 m / s )
7.39
λ =
7.40
Step 1: The equation needed to solve this problem is the De Broglie equation.
λ =
h
mu
Substitute the known values into the equation to solve for the wavelength of the proton.
λ =
h
=
mu
F 1 kg ⋅ m / s I
GH 1 J JK
−
= 1.37 × 10 m
2
(6.63 × 10 −34 J ⋅ s) ×
2
15
× 10
(1673
.
−27
8
kg)(2.90 × 10 m / s)
Step 2: Convert wavelength in units of meters to nanometers.
λ = 1.37 × 10
7.41
−15
m ×
1 nm
1 × 10
−9
m
= 1.37 × 10
−6
nm
Converting the velocity to units of m/s::
 1.20 × 102 mi   1.61 km   1000 m  1 hr 


 = 53.7 m / s

  1 mi   1 km 
1h
 3600 s 


λ=
7.42
h
6.63 × 10 −34 J ⋅ s
=
= 9.96 × 10 −34 m = 9.96 × 10 −32 cm
mu ( 0.0124 kg )(53.7 m / s )
First, we convert mph to m/s.
35 mi
1.61 km
1000 m
1 hr
×
×
×
= 16 m/s
1 hr
1 mi
1 km
3600 s
λ =
h
=
mu
(6.63 × 10 −34 J ⋅ s) ×
F 1 kg ⋅ m / s I
GH 1 J JK
−
−
= 1.7 × 10 m = 1.7 × 10 nm
2
2
32
(2.5 × 10
−3
kg)(16 m / s)
23
125
126
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
7.53
The angular momentum quantum number l can have integral (i.e. whole number) values from 0 to n−1. In
this case n = 2, so the allowed values of the angular momentum quantum number are 0 and 1.
Each allowed value of the angular momentum quantum number labels a subshell. Within a given subshell
(label l) there are 2l + 1 allowed energy states (orbitals) each labeled by a different value of the magnetic
quantum number. The allowed values run from −l through 0 to +l (whole numbers only). For the subshell
labeled by the angular momentum quantum number l = 1, the allowed values of the magnetic quantum number
are −1, 0, and 1. For the other subshell in this problem labeled by the angular momentum quantum number
l = 0, the allowed value of the magnetic quantum number is 0.
If the allowed whole number values run from −1 to +1, are there always 2l + 1 values? Why?
7.54
When n = 3, there are three possible values of l: 0, 1, and 2, corresponding to the s, p, and d orbitals,
respectively. The values of ml for each l value are:
l = 0:
7.55
(a)
(b)
(c)
ml = 0
l = 1:
ml = −1, 0, 1
l = 2:
ml = −2, −1, 0, 1, 2
2p: n = 2, l = 1, ml = 1, 0, or −1
3s: n = 3, l = 0, ml = 0 (only allowed value)
5d: n = 5, l = 2, ml = 2, 1, 0, −1, or −2
An orbital in a subshell can have any of the allowed values of the magnetic quantum number for that
subshell. All the orbitals in a subshell have exactly the same energy.
7.56
(a)
The number given in the designation of the subshell is the principal quantum number, so in this case
n = 4. For p orbitals, l = 1. ml can have integer values from −l to +l. Therefore, ml can be −1, 0, and +1.
The three values for ml correspond to the three p orbitals.
Following the same reasoning as part (a)
(b)
3d: n = 3, l = 2, ml = −2, −1, 0, 1, or 2 (5 orbitals)
(c)
3s: n = 3, l = 0, ml = 0 (1 orbital)
(d)
5f: n = 5, l = 3, ml = −3, −2, −1, 0, 1, 2, or 3 (7 orbitals)
7.57
A 2s orbital is larger than a 1s orbital. Both have the same spherical shape. The 1s orbital is lower in energy
than the 2s.
7.58
The two orbitals are identical in size, shape, and energy. They differ only in their orientation with respect to
each other.
Can you assign a specific value of the magnetic quantum number to these orbitals? What are the allowed
values of the magnetic quantum number for the 2p subshell?
7.59
The allowed values of l are 0, 1, 2, 3, and 4. These correspond to the 5s, 5p, 5d, 5f, and 5g subshells. These
subshells each has one, three, five, seven, and nine orbitals, respectively.
7.60
For n = 6, the allowed values of l are 0, 1, 2, 3, 4, and 5 [l = 0, ..., (n − 1), integer values]. These l values
correspond to the 6s, 6p, 6d, 6f, 6g, and 6h subshells. These subshells each have 1, 3, 5, 7, 9, and 11 orbitals,
respectively (number of orbitals = 2l + 1).
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
7.61
127
There can be a maximum of two electrons occupying one orbital.
(a)
two;
six;
(b)
(c)
ten;
(d)
fourteen.
What rule of nature demands a maximum of two electrons per orbital? Do they have the same energy? How
are they different? Would five 4d orbitals hold as many electrons as five 3d orbitals? In other words, does
the principal quantum number n affect the number of electrons in a given subshell?
7.62
n value
1
2
3
4
5
6
orbital sum
1
1+3=4
1+3+5=9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
1 + 3 + 5 + 7 + 9 + 11 = 36
total number of electrons
2
8
18
32
50
72
2
In each case the total number of orbitals is just the square of the n value (n ). The total number of electrons
2
is 2n .
7.63
3s: two;
3d: ten;
4p: six;
4f: fourteen;
7.64
The electron configurations for the elements are
5f: fourteen
(a)
N: 1s 2s 2p
2
2
3
(b)
Si: 1s 2s 2p 3s 3p
2
2
6
2
2
There are six s-type electrons.
(c)
S: 1s 2s 2p 3s 3p
2
2
6
2
4
There are no d-type electrons.
There are three p-type electrons.
7.65
See Figure 7.18 in your textbook.
7.66
In the many-electron atom, the 3p orbital electrons are more effectively shielded by the inner electrons of the
atom (that is, the 1s, 2s, and 2p electrons) than the 3s electrons. The 3s orbital is said to be more
“penetrating” than the 3p and 3d orbitals. In the hydrogen atom there is only one electron, so the 3s, 3p, and
3d orbitals have the same energy.
7.67
Equation (7.4) of the text gives the orbital energy terms of the principal quantum number, n, alone (for the
hydrogen atom). The energy does not depend on any of the other quantum numbers. If two orbitals in the
hydrogen atom have the same value of n, they have equal energy.
(a)
2s > 1s;
(b)
3p > 2p;
equal;
(c) 3s < 4s
(d)
equal;
(e)
5s > 4f.
7.68
(a) 2s < 2p
7.77
(a)
is wrong because the magnetic quantum number ml can have only whole number values.
(c)
is wrong because the maximum value of the angular momentum quantum number l is n − 1.
(e)
is wrong because the electron spin quantum number ms can have only half−integral values.
7.78
(b) 3p < 3d
(c)
(d) 4d < 5f
For aluminum, there are not enough electrons in the 2p subshell. (The 2p subshell holds six electrons.) The
2 2 6 2 1
number of electrons (13) is correct. The electron configuration should be 1s 2s 2p 3s 3p . The
configuration shown might be an excited state of an aluminum atom.
128
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
For boron, there are too many electrons. (Boron only has five electrons.) The electron configuration should
2 2 1
be 1s 2s 2p . What would be the electric charge of a boron ion with the electron arrangement given in the
problem?
For fluorine, there are also too many electrons. (Fluorine only has nine electrons.) The configuration shown
−
2 2 5
is that of the F ion. The correct electron configuration is 1s 2s 2p .
7.79
Since the atomic number is odd, it is mathematically impossible for all the electrons to be paired. There must
be at least one that is unpaired. The element would be paramagnetic.
7.80
You should write the electron configurations for each of these elements to answer this question. In some
cases, an orbital diagram may be helpful.
2
1
B: [He]2s 2p (1 unpaired electron)
2 3
P: [Ne]3s 3p (3 unpaired electrons)
2 5
Mn: [Ar]4s 3d (5 unpaired electrons)
2 2
Zr: [Kr]5s 4d (2 unpaired electrons)
2 10
Cd: [Kr]5s 4d (0 unpaired electrons)
2 14 4
W: [Xe]6s 4f 5d (4 unpaired electrons)
2 1 1
Ce: [Xe]6s 5d 4f (2 unpaired electrons)
7.81
2
2
1
B:
1s 2s 2p
V:
[Ar]4s 3d
Ni:
[Ar]4s 3d
Ne:
Sc:
Se:
Ru:
I:
Pb:
Ho:
(0 unpaired electrons, Why?)
2 1
[Ar]4s 3d (1 unpaired electron)
2 10 4
[Ar]4s 3d 4p (2 unpaired electrons)
1 7
[Kr]5s 4d (4 unpaired electrons)
2 10 5
[Kr]5s 4d 5p (1 unpaired electron)
2 14 10 2
[Xe]6s 4f 5d 6p (2 unpaired electrons)
2 11
[Xe]6s 4f (3 unpaired electrons)
As:
[Ar]4s 3d 4p
2
10
3
2
10
5
1
14
10
2
3
I:
[Kr]5s 4d 5p
2
8
Au:
[Xe]6s 4f 5d
What is the meaning of “[Ar]”? of “[Kr]”? of “[Xe]”?
7.82
Step 1: Germanium (Ge) has 32 electrons. These electrons need to be placed in atomic orbitals.
Step 2: The noble gas element that most nearly proceeds Se is Ar. Therefore, the noble gas core is [Ar].
This core accounts for 18 electrons.
Step 3: See Figure 7.24 of your text to check the order of filling subshells past the Ar noble gas core. You
should find that the order of filling is 4s, 3d, then 4p. There are 14 remaining electrons to distribute
among these orbitals.
The 4s orbital can hold 2 electrons. Each of the five 3d orbitals can hold 2 electrons for a total of 10
electrons. This leaves 2 electrons to fill the 4p orbitals.
The electrons configuration for Ge is:
2
10
[Ar] 4s 3d 4p
2
You should follow the same reasoning for the remaining atoms.
2
6
Fe: [Ar]4s 3d
2 14 4
W: [Xe]6s 4f 5d
2
10
Zn: [Ar]4s 3d
2 14 10 1
Tl: [Xe]6s 4f 5d 6p
1
Ru: [Kr]5s 4d
7
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
7.83
There are a total of twelve electrons:
Orbital
n
l
ml
ms
1
+
2
1
−
2
1
+
2
1
−
2
1
+
2
1
−
2
1
+
2
1
−
2
1
+
2
1
−
2
1
+
2
1
−
2
1s
1
0
0
1s
1
0
0
2s
2
0
0
2s
2
0
0
2p
2
1
1
2p
2
1
1
2p
2
1
0
2p
2
1
0
2p
2
1
−1
2p
2
1
−1
3s
3
0
0
3s
3
0
0
↑↓
2
3s
↑↓ ↑↓ ↑
5
3p
The element is magnesium.
7.84
↑↓
2
3s
↑
↑ ↑
3
3p
+
S (5 valence electrons)
3 unpaired electrons
↑↓
2
3s
↑↓ ↑ ↑
4
3p
S (6 valence electrons)
2 unpaired electrons
−
S (7 valence electrons)
1 unpaired electron
+
S has the most unpaired electrons
7.85
We first calculate the wavelength, then we find the color using Figure 7.2 of the text.
λ=
hc ( 6.63 × 10 −34 J ⋅ s )(3.00 × 108 m / s )
=
= 4.63 × 10 −7 m or 463 nm, which is blue.
−19
E
4.30 × 10
J
7.86
Part (b) is correct in the view of contemporary quantum theory. Bohr’s explanation of emission and
absorption line spectra appears to have universal validity. Parts (a) and (c) are artifacts of Bohr’s early
planetary model of the hydrogen atom and are not considered to be valid today.
7.87
(a)
Wavelength and frequency are reciprocally related properties of any wave. The two are connected
through Equation (7.1) of the text. See Example 7.1 of the text for a simple application of the
relationship to a light wave.
129
130
7.88
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
(b)
Typical wave properties: wavelength, frequency, characteristic wave speed (sound, light, etc.). Typical
particle properties: mass, speed or velocity, momentum (mass × velocity), kinetic energy. For
phenomena that we normally perceive in everyday life (macroscopic world) these properties are
mutually exclusive. At the atomic level (microscopic world) objects can exhibit characteristic
properties of both particles and waves. This is completely outside the realm of our everyday common
sense experience and is extremely difficult to visualize.
(c)
Quantization of energy means that emission or absorption of only descrete energies is allowed
(e.g., atomic line spectra). Continuous variation in energy means that all energy changes are allowed
(e.g., continuous spectra).
(a)
With n = 2, there are n orbitals = 2 = 4. ms = +1/2, specifies 1 electron per orbital, for a total of
4 electrons.
(b)
n = 4 and ml = +1, specifies one orbital in each subshell with l = 1, 2, or 3 (i.e., a 4p, 4d, and 4f orbital).
Each of the three orbitals holds 2 electrons for a total of 6 electrons.
(c)
If n = 3 and l = 2, ml has the values 2, 1, 0, −1, or −2. Each of the five orbitals can hold 2 electrons for
−
a total of 10 electrons (2 e in each of the five 3d orbitals).
(d)
If n = 2 and l = 0, then ml can only be zero. ms = −1/2 specifies 1 electron in this orbital for a total of
−
1 electron (one e in the 2s orbital).
(e)
n = 4, l = 3 and ml = −2, specifies one 4f orbital. This orbital can hold 2 electrons.
2
2
7.89
See the appropriate sections of the textbook in Chapter 7.
7.90
The wave properties of electrons are used in the operation of an electron microscope.
7.91
In the photoelectric effect, light of sufficient energy shining on a metal surface causes electrons to be ejected
(photoelectrons). Since the electrons are charged particles, the metal surface becomes positively charged as
more electrons are lost. After a long enough period of time, the positive surface charge becomes large
enough to start attracting the ejected electrons back toward the metal with the result that the kinetic energy of
the departing electrons becomes smaller.
7.92
(a)
First convert 100 mph to units of m/s.
100 mi
1 hr
1.609 km
1000 m
×
×
×
= 44.7 m/s
1 hr
3600 s
1 mi
1 km
Using the de Broglie equation:
λ =
(b)
h
=
mu
( 6.63 × 10 −34 J ⋅ s ) ×
F 1 kg ⋅ m / s I
GH 1 J JK
= 1.05 × 10
( 0141
.
kg )( 44.7 m / s )
2
2
−34
m = 1.05 × 10
−25
The average mass of a hydrogen atom is:
1.008
1 mol
g
−24
−27
×
= 1.674 × 10 g/H atom = 1.674 × 10 kg
mol
6.022 × 1023 atoms
nm
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
131
 1 kg ⋅ m 2 / s 2 
(6.63 × 10−34 J ⋅ s) × 



1J
h
−9


λ =
=
= 8.86 × 10 m = 8.86 nm
−
27
mu
(1.674 × 10
kg)(44.7 m / s)
7.93
There are many more paramagnetic atoms than diamagnetic atoms because of Hund's rule.
7.94
First, let’s find the energy needed to photodissociate one water molecule.
2858
. kJ
1 mol
−22
−19
= 4.746 × 10 kJ/molecule = 4.746 × 10 J/molecule
×
1 mol
6.022 × 1023 molecules
The maximum wavelength of a photon that would provide the above energy is:
λ =
(6.63 × 10 −34 J ⋅ s)(3.00 × 108 m / s)
hc
−7
=
= 4.19 × 10 m = 419 nm
E
4.746 × 10 −19 J
This wavelength is in the visible region of the electromagnetic spectrum. Since water is continuously being
struck by visible radiation without decomposition, it seems unlikely that photodissociation of water by this
method is feasible.
7.95
For the Lyman series, we want the longest wavelength (smallest energy), with ni = 2 and nf = 1. Using
Equation (7.5) of the text:
∆E = RH
λ =
F 1 − 1 I = (218
. × 10 18 J
GH ni2 nf 2 JK . × 10 18 J )FGH 212 − 112 IJK = − 164
−
−
hc
( 6.63 × 10 −34 J ⋅ s )(3.00 × 108 m / s )
−7
=
= 1.21 × 10 m = 121 nm
∆E
164
. × 10 −18 J
For the Balmer series, we want the shortest wavelength (highest energy), with ni = ∞ and nf = 2.
∆E = RH
λ =
F 1 − 1 I = (218
GH ni2 nf 2 JK . × 10 18 J )FGH ∞12 − 212 IJK = − 5.45 × 10 19 J
−
−
hc
( 6.63 × 10 −34 J ⋅ s )(3.00 × 108 m / s )
−7
=
= 3.65 × 10 m = 365 nm
−
19
∆E
5.45 × 10
J
Therefore the two series do not overlap.
7.96
The Balmer series corresponds to transitions to the n = 2 level.
+
For He :
∆E = RHe +
F 1 − 1I
GH ni2 nf2 JK
λ=
(6.63 × 10 −34 J ⋅ s)(3.00 × 108 m / s)
hc
=
∆E
∆E
132
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
For the transition, n = 3 → 2
∆E = (8.72 × 10
−18
J)
FG 1 − 1 IJ = −1.21 × 10
H3 2 K
2
−18
2
For the transition, n = 4 → 2, ∆E = −1.64 × 10
−18
For the transition, n = 5 → 2, ∆E = −1.83 × 10
−18
For the transition, n = 6 → 2, ∆E = −1.94 × 10
−18
J
λ=
199
. × 10 −25 J ⋅ m
121
. × 10 −18 J
J
λ = 121 nm
J
λ = 109 nm
J
λ = 103 nm
= 1.64 × 10
−7
m = 164 nm
For H, the calculations are identical to those above, except the Rydberg constant for H is 2.18 × 10
For the transition, n = 3 → 2, ∆E = −3.03 × 10
−19
For the transition, n = 4 → 2, ∆E = −4.09 × 10
−19
For the transition, n = 5 → 2, ∆E = −4.58 × 10
−19
For the transition, n = 6 → 2, ∆E = −4.84 × 10
−19
J
λ = 657 nm
J
λ = 487 nm
J
λ = 434 nm
J
λ = 411 nm
−18
J.
+
All the Balmer transitions for He are in the ultraviolet region; whereas, the transitions for H are all in the
visible region. Note the negative sign for energy indicating that a photon has been emitted.
7.97
(a)
∆H° = ∆H fD (O) + ∆H fD (O2) − ∆H fD (O3) = 249.4 kJ + 0 kJ − 142.2 kJ = 107.2 kJ
(b)
The energy in part (a) is for one mole of photons. To apply E = hν we must divide by Avogadro’s
number. The energy of one photon is:
E =
FG 107.2 kJ IJ F 1 mol I FG 1000 J IJ = 1.780 × 10
H 1 mol K GH 6.022 × 10 JK H 1 kJ K
23
−19
J
The wavelength of this photon can be found using the relationship E = hc/λ.
λ =
7.98
F
GH
hc
( 6.63 × 10 −34 J ⋅ s )(3.00 × 108 m / s )
109 nm
=
×
E
1m
× 10 −19 J
1780
.
I = 1.12 × 10 nm
JK
3
Since the energy corresponding to a photon of wavelength λ1 equals the energy of photon of wavelength λ2
plus the energy of photon of wavelength λ3, then the equation must relate the wavelength to energy.
energy of photon 1 = (energy of photon 2 + energy of photon 3)
Since E =
hc
, then:
λ
hc
hc
hc
=
+
λ1
λ2
λ3
Dividing by hc:
1
1
1
=
+
λ1
λ2
λ3
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
7.99
133
The difference in ionization energy is:
(412 − 126)kJ/mol = 286 kJ/mol.
In terms of one atom:
286 × 103 J
1 mol
−19
= 4.75 × 10 J/atom
×
23
1 mol
6.022 × 10 atoms
λ =
7.100
hc
( 6.63 × 10 −34 J ⋅ s )(3.00 × 108 m / s )
−7
=
= 4.19 × 10 m = 419 nm
∆E
4.75 × 10 −19 J
We use Heisenberg’s uncertainty principle with the equality sign to calculate the minimum uncertainty.
∆x∆p =
h
4π
The momentum (p) is equal to the mass times the velocity.
p = mu
or
∆p = m∆u
We can write:
∆p = m∆u =
h
4π∆x
Finally, the uncertainty in the velocity of the oxygen molecule is:
∆u =
7.101
(6.63 × 10 −34 J ⋅ s)
h
−5
= 2.0 × 10 m/s
=
−26
−5
4πm∆x
4π (5.3 × 10
kg)(5.0 × 10 m)
It takes:
2
(5.0 × 10 g ice) ×
334 J
5
2
= 1.67 × 10 J to melt 5.0 × 10 g of ice.
1 g ice
Energy of a photon with a wavelength of 660 nm:
E =
( 6.63 × 10 −34 J ⋅ s )(3.00 × 108 m / s )
hc
−19
=
= 3.01 × 10 J
λ
660 × 10 −9 m
2
Number of photons needed to melt 5.0 × 10 g of ice:
(167
. × 105 J ) ×
1 photon
3.01 × 10
−19
J
= 5.5 × 10
23
photons
The number of water molecules is:
1 mol H 2O
6.022 × 1023 H 2O molecules
2
25
×
= 1.7 × 10 H2O molecules
(5.0 × 10 g H2O) ×
18.02 g H 2 O
1 mol H 2O
134
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
The number of water molecules converted from ice to water by one photon is:
17
. × 1025 H 2O molecules
5.5 × 1023 photons
7.102
= 31 H2O molecules/photon
13+
As an estimate, we can equate the energy for ionization (Fe
(3/2RT) of the ions.
14+
→ Fe
) to the average kinetic energy
35
. × 104 kJ
1000 J
7
×
= 3.5 × 10 J
1 mol
1 kJ
IE =
3
RT
2
7
3.5 × 10 J/mol =
3
(8.314 J / mol ⋅ K)T
2
6
T = 2.8 × 10 K
The actual temperature can be, and most probably is, higher than this.
7.103
Energy of a photon at 360 nm:
Ε = hν =
hc
(6.63 × 10−34 J ⋅ s)(3.00 × 108 m / s)
−19
=
= 5.53 × 10 J
−
9
λ
360 × 10 m
2
Area of exposed body in cm :
 1 cm 
0.45 m × 

 1 × 10−2 m 


2
2
3
2
= 4.5 × 10 cm
The number of photons absorbed by the body in 2 hours is:
 2.0 × 1016 photons 
 7200 s 
23
0.5 × 
 × (4.5 × 103 cm 2 ) × 
 = 3.2 × 10 photons / 2 hr
2


2
hr
cm ⋅ s




The factor of 0.5 is used above because only 50% of the radiation is absorbed.
3.2 × 10
23
photons with a wavelength of 360 nm correspond to an energy of:
 5.53 × 10−19 J 
(3.2 × 1023 photons) × 
 = 1.8 × 105 J
 1 photon 


7.104
A “blue” photon (shorter wavelength) is higher energy than a “yellow” photon. For the same amount of
energy delivered to the metal surface, there must be fewer “blue” photons than “yellow” photons. Thus, the
yellow light would eject more electrons since there are more “yellow” photons. Since the “blue” photons are
of higher energy, blue light will eject electrons with greater kinetic energy.
CHAPTER 7: THE ELECTRONIC STRUCTURE OF ATOMS
7.105
135
h
, the mass of an N2 molecule (in kg) and the velocity of an N2
mu
molecule (in m/s) is needed to calculate the deBroglie wavelength of N2.
Looking at the deBroglie equation λ =
First, calculate the root-mean-square velocity of N2.
M(N2) = 28.02 g/mol = 0.02802 kg/mol
urms (N 2 ) =
J 

(3)  8.314
( 300 K )
mol ⋅ K 

= 516.8 m / s
kg 

0.02802

mol 

Second, calculate the mass of one N2 molecule in kilograms.
28.02 g N 2
1 mol N 2
1 kg
×
×
= 4.653 × 10−26 kg / molecule
23
1 mol N 2
6.022 × 10 N 2 molecules 1000 g
Now, substitute the mass of an N2 molecule and the root-mean-square velocity into the deBroglie equation to
solve for the deBroglie wavelength of an N2 molecule.
λ=
h
(6.626 × 10−34 J ⋅ s)
=
= 2.755 × 10−11 m
mu (4.653 × 10−26 kg)(516.8 m / s)
7.106
See appropriate sections of the text. Planck's quantum theory is discussed in Section 7.1, the photoelectric
effect is discussed in Section 7.2, Bohr's model of the hydrogen atom is discussed in Section 7.3, de Broglie's
particle-wave relation is discussed in Section 7.4, and Heisenberg's uncertainty principle is discussed in
Section 7.5.
7.107
(a)
False. n = 2 is the first excited state.
(b)
False. In the n = 4 state, the electron is (on average) further from the nucleus and hence easier to
remove.
(c)
True.
(d)
False. The n = 4 to n = 1 transition is a higher energy transition, which corresponds to a shorter
wavelength.
(e)
True.
7.108
Based on the selection rule, which states that ∆l = ±1, only (b) and (d) are allowed transitions.